HDU5492-Find a path_服务器 -bash: sodu: 未找到命令-优快云博客

解决一个迷宫问题，从起点到终点寻找一条路径，使得路径上的数字序列具有最小的方差值，通过动态规划实现。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

Find a path

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1701 Accepted Submission(s): 738

Problem Description

Frog fell into a maze. This maze is a rectangle containing

N rows and

M columns. Each grid in this maze contains a number, which is called the magic value. Frog now stays at grid (1, 1), and he wants to go to grid (N, M). For each step, he can go to either the grid right to his current location or the grid below his location. Formally, he can move from grid (x, y) to (x + 1, y) or (x, y +1), if the grid he wants to go exists.
Frog is a perfectionist, so he'd like to find the most beautiful path. He defines the beauty of a path in the following way. Let’s denote the magic values along a path from (1, 1) to (n, m) as

A1,A2,…AN+M−1 , and

Aavg is the average value of all

Ai . The beauty of the path is

(N+M–1) multiplies the variance of the values:

(N+M−1)∑N+M−1i=1(Ai−Aavg)2
In Frog's opinion, the smaller, the better. A path with smaller beauty value is more beautiful. He asks you to help him find the most beautiful path.

Input

The first line of input contains a number

T indicating the number of test cases (

T≤50 ).
Each test case starts with a line containing two integers

N and

M (

1≤N,M≤30 ). Each of the next

N lines contains

M non-negative integers, indicating the magic values. The magic values are no greater than 30.

Output

For each test case, output a single line consisting of “Case #X: Y”.

X is the test case number starting from 1.

Y is the minimum beauty value.

Sample Input

Sample Output

  
   Case #1: 14

Source

2015 ACM/ICPC Asia Regional Hefei Online

Recommend

wange2014

题意：给N*M(1<=N，M<=30)的矩阵，矩阵的每一格有一个非负权值(<=30)，从(1，1)出发，每次只能向右或向下移动，到达(n，m)时，经过的格子的权值形成序列A，求(n + m - 1)*sigma(Ai - Aavg)的最小值，Aavg是路径上Ai的平均数

解题思路：将式子展开后，化简整理可得：(N+M-1)*s1-s2。其中s1是序列A的平方和，s2是序列A的和的平方。因为序列A的和不会超过(30+30-1)*30。设dp[i][j][k]表示到达(i，j)，序列和为k时，序列的平方和的最小值。

#include <iostream>
#include <cstdio>
#include <cstring>
#include <string>
#include <algorithm>
#include <cmath>
#include <map>
#include <set>
#include <stack>
#include <queue>
#include <vector>
#include <bitset>
#include <functional>

using namespace std;

#define LL long long
const int INF = 0x3f3f3f3f;

int a[39][39];
int dp[39][39][2000];
int n, m;

int main()
{
	int t,cas=0;
	scanf("%d", &t);
	while (t--)
	{
		scanf("%d%d", &n, &m);
		for (int i = 1; i <= n; i++)
			for (int j = 1; j <= m; j++)
				scanf("%d", &a[i][j]);
		memset(dp, INF, sizeof dp);
		dp[1][1][a[1][1]] = a[1][1] * a[1][1];
		for (int i = 1; i <= n; i++)
		{
			for (int j = 1; j <= m; j++)
			{
				if (i == 1 && j == 1) continue;
				for (int k = 0; k <= 1900; k++)
				{
					if (i >= 2 && k >= a[i][j]) dp[i][j][k] = min(dp[i][j][k], dp[i - 1][j][k - a[i][j]] + a[i][j] * a[i][j]);
					if (j >= 2 && k >= a[i][j]) dp[i][j][k] = min(dp[i][j][k], dp[i][j - 1][k - a[i][j]] + a[i][j] * a[i][j]);
				}
			}
		}
		int ans = INF;
		for (int i = 0; i <= 1900; i++)
		{
			if (dp[n][m][i] == INF) continue;
			ans =min(ans, (n + m - 1)*dp[n][m][i] - i*i);
		}
		printf("Case #%d: %d\n", ++cas, ans);
	}
	return 0;
}